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olasank [31]
3 years ago
14

under what conditions would a scale record a weight for an object which is equal to the magnitude of the gravitational force on

the object?
Physics
2 answers:
pentagon [3]3 years ago
7 0

answer:if there are no external forces acting on the object

Explanation:

under what conditions would a scale record a weight for an object which is equal to the magnitude of the gravitational force on the object?

If gravity and the normal reaction force are the only forces acting on the body, and the body is at rest or moving at constant velocity, the normal force and the weight are equal and opposite. Also if there are no influence of external forces

and air resistance is negligible.

from newtons law of motion which states that the rate of change in momentum is directly proportional to the force applied

f=km(v-u)/t

f=kma

where k is the constant and  equal to 1

f=ma

d1i1m1o1n [39]3 years ago
4 0
In a non accelerated frame, also called inertial frame, when no additional forces are acting upon the body.

For instance, a scale in free fall will not record the weight, but zero, like the astronauts. But also, if someone is pulling the object or pushing it, the scale will record some other value.
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Two objects (42.0 and 21.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley h
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Answer:

a = 3.27 m/s²

T = 275 N

Explanation:

Given that:

Mass m₁ = 42.p0 kg

Mass m₂ = 21.0 kg

Consider both masses to be in a whole system, then:

The acceleration can be determined as:

(m_1+m_2)a = g(m_1-m_2)

Making acceleration the subject in the above formula;

a =\dfrac{g(m_1-m_2)}{(m_1+m_2)}

a =\dfrac{9.8(42.0-21.0)}{(42.0+21.0)}

a =\dfrac{9.8(21.0)}{(63.0)}

a =\dfrac{205.8}{(63.0)}

a = 3.27 m/s²

in the string, the tension is calculated using the formula:

T = \dfrac{2m_1m_2g}{(m_1+m_2)}

T = \dfrac{2(42)(21)(9.81)}{(42+21)}

T = \dfrac{17304.84}{63}

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8 0
2 years ago
At an instant when the displacement is equal to a/2, what fraction of the total energy of the system is potential?
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A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences
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Answer:

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Explanation:

It is given that,

Charge, q=3\times 10^{-6}\ C

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Acceleration, a=3\times 10^{4}\ m/s^2

We need to find the minimum magnetic field that would produce such an acceleration. So,

ma=qvB\ sin\theta

For minimum magnetic field,

ma=qvB

B=\dfrac{ma}{qv}

B=\dfrac{2\times 10^{-6}\ C\times 3\times 10^{4}\ m/s^2}{3\times 10^{-6}\ C\times 5\times 10^{6}\ m/s}

B = 0.004 T

or

B = 4 mT

So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.

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3 years ago
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